Suppose that
Radius of the base is r, and the height of the original cone is h.
Then,
Volume of the original cone (V1)=13πr2h
When cone reduced then,
radius =r2
height =h
Now,
Volume of the reduced cone (V2)=13π(r2)2h
=14(13πr2h)
=14V1$
∴V2V1=14=1:4
This is the required solution.